Simplify and expand the following expression: $ \dfrac{2}{3t - 3}- \dfrac{4}{2t - 14}- \dfrac{2t}{t^2 - 8t + 7} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{2}{3t - 3} = \dfrac{2}{3(t - 1)}$ We can factor a $2$ out of denominator in the second term: $ \dfrac{4}{2t - 14} = \dfrac{4}{2(t - 7)}$ We can factor the quadratic in the third term: $ \dfrac{2t}{t^2 - 8t + 7} = \dfrac{2t}{(t - 1)(t - 7)}$ Now we have: $ \dfrac{2}{3(t - 1)}- \dfrac{4}{2(t - 7)}- \dfrac{2t}{(t - 1)(t - 7)} $ The least common multiple of the denominators is: $ 6(t - 1)(t - 7)$ In order to get the first term over $6(t - 1)(t - 7)$ , multiply by $\dfrac{2(t - 7)}{2(t - 7)}$ $ \dfrac{2}{3(t - 1)} \times \dfrac{2(t - 7)}{2(t - 7)} = \dfrac{4(t - 7)}{6(t - 1)(t - 7)} $ In order to get the second term over $6(t - 1)(t - 7)$ , multiply by $\dfrac{3(t - 1)}{3(t - 1)}$ $ \dfrac{4}{2(t - 7)} \times \dfrac{3(t - 1)}{3(t - 1)} = \dfrac{12(t - 1)}{6(t - 1)(t - 7)} $ In order to get the third term over $6(t - 1)(t - 7)$ , multiply by $\dfrac{6}{6}$ $ \dfrac{2t}{(t - 1)(t - 7)} \times \dfrac{6}{6} = \dfrac{12t}{6(t - 1)(t - 7)} $ Now we have: $ \dfrac{4(t - 7)}{6(t - 1)(t - 7)} - \dfrac{12(t - 1)}{6(t - 1)(t - 7)} - \dfrac{12t}{6(t - 1)(t - 7)} $ $ = \dfrac{ 4(t - 7) - 12(t - 1) - 12t} {6(t - 1)(t - 7)} $ Expand: $ = \dfrac{4t - 28 - 12t + 12 - 12t}{6t^2 - 48t + 42} $ $ = \dfrac{-20t - 16}{6t^2 - 48t + 42}$ Simplify: $ = \dfrac{-10t - 8}{3t^2 - 24t + 21}$